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Enthalpy of water at triple point and freezing point

  1. Apr 28, 2014 #1
    By definition we have h=u+pv for a system

    where h=specific enthalpy
    u=specific internal energy
    p=external pressure applied on the system
    v=specific volume of the system

    The internal energy is arbitrarily taken to be zero at the triple point of water i.e, u=0 when T=273.16K and P=4.587 mm of Hg.

    So at the triple point of water we have enthalpy of water to be slightly positive due to small value of the 'pv' term.

    At freezing point,however,we have T=273.16K and P=760 mm of Hg.

    So the change in enthalpy could be obtained as Δh=Δu+Δ(pv)

    Now, Δu=0 as ΔT=0

    and the only difference in enthalpy is due to the Δ(pv) .

    At triple point, we have a range of values for 'v' along the triple point line and p=4.587 mm of Hg.

    At freezing point, we have v=0.001 m3/kg and p=760 mm of Hg.

    So to bring a unit mass of water from triple point to the freezing point a reversible isothermal path could be followed at T=273.16K from P1=4.587 mm of Hg to P2=760 mm of Hg.

    Thus the change in enthalpy would be equal to the displacement work done to bring water from the triple point to the freezing point i.e, Δ(pv).

    But in most text books the enthalpy of water at both triple and freezing points are taken to be equal.Whereas they cannot be theoretically equal in value is this because the change in specific volume is too small to take Δ(pv) into account?
     
  2. jcsd
  3. Apr 28, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Liquid water is not an ideal gas.
     
  4. Apr 28, 2014 #3
    Somalya,

    You are saying that the internal energy of liquid water at the triple point is taken as zero, correct? Then, yes, the enthalpy of liquid water at 273.16 K and 4.587 mm Hg is pv, where v is the specific volume of liquid water at 273.16 K and 4.587 mm Hg, and p is 4.587 mm Hg. For liquid water at 273.16 K and 760 mm Hg, the enthalpy is pv, where v is the specific volume of liquid water at 273.16 and 760 mm Hg, and p is 760 mm Hg. Of course, in both these cases, you have to use appropriate units. The specific volume of liquid water does not change much between 4.587 mm Hg and 760 mm Hg. So the change in enthalpy between the two cases is basically vΔp, where Δp=(760 - 4.587) mm Hg. Again, you have to use consistent units. Also, since the specific volume of the liquid is very low (1cc/gm), the enthalpy in either case is very low.

    Chet
     
  5. Apr 29, 2014 #4
    Thank You Chet

    That's what I was confused about..........so the enthalpy is pretty much taken the same and closely equal to zero.
     
  6. Apr 29, 2014 #5
    Am I wrong here?

    I mean as far as I know,Δu=0 when ΔT=0 regardless of the working substance being an ideal gas or any other non-ideal substance.

    Is it so that my assumption is incorrect for liquid water?
     
  7. Apr 29, 2014 #6
    Yes. Calculate it yourself and see what you get. It should only take a minute or two. Compare what you get with what is in your table.

    Chet
     
    Last edited: Apr 29, 2014
  8. Apr 29, 2014 #7
    In general, u is a function of both temperature and pressure. However, for the situation you are looking at, the effect of pressure on u is orders of magnitude lower than the effect of pressure on h (because h has that pv term). For gases at pressures beyond the ideal gas range, the effect of pressure on both u and h need to be taken into account.

    Chet
     
  9. Apr 30, 2014 #8
    I just realized it might be wrong to calculate enthalpy the way previously assumed.

    Enthalpy is the heat added at constant pressure Δh=CpdT+pΔv.

    To take the system at the triple point of water (at a low pressure) to it's freezing point(at a higher pressure) we are following an isothermal path and not an isobaric path so heat added during the process is not heat added at constant pressure.

    But the internal energy at the triple point of water is assigned a value zero and hence having practically zero enthalpy as the value of the 'pv' term is almost negligible.So if I wish to calculate the change in enthalpy for water at it's freezing point(at a higher pressure)taking the value of enthalpy at the triple point of water to be zero we must elevate the system at the triple point to a higher pressure at the freezing point and hence a constant pressure process cannot be followed!

    Then how is the value of enthalpy for a system at higher pressures assigned?
     
  10. Apr 30, 2014 #9
    internal energy of water at 273.16 and 4.587 mm Hg = 0

    4.187 mm Hg = (133.32)(4.587)=611.5 Pa=611.5 N/m2
    specific volume of water ~ 0.00100 m3/kg

    pv = (0.001m3/kg)(611.5 N/m2)=0.612 Nm/kg=0.612 J/kg

    For p = 760 mm Hg, pv = 101.3 J/kg

    enthalpy at 273.16 and 4.587 mm Hg = 0.612 J/kg

    enthaply at 273.16 and 760 mm Hg = 101.3 J/kg

    How do these value compare with those in your table?

    Chet
     
    Last edited: Apr 30, 2014
  11. May 1, 2014 #10
    Well couldn't find the values at such low pressures as triple point:frown:
     
  12. May 1, 2014 #11
    What about the point at 273 and 760?

    Chet
     
  13. May 2, 2014 #12
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